Chapter 20, Problem 20.11P

(a)

Interpretation Introduction

Interpretation:

The resolution is required for grating to resolve given wavelengths has to be calculated.

Concept introduction:

Resolution of grating:

It is useful for separate two closely merged peaks

$\begin{array}{l}\frac{\text{λ}}{\text{Δλ}}\text{=nN}\\ \text{where,}\\ \text{λ}\text{is wavelength}\\ \text{n }\text{is the diffraction}\text{order}\\ \text{N}\text{is the number of grooves of the grating}\end{array}$

Answer to Problem 20.11P

The resolution is $\text{17066}\text{.666}$

Explanation of Solution

Given,

Wavelengths is $\text{512}\text{.23}\text{and 512}\text{.26}\text{nm}$

According to resolution of grating:

$\begin{array}{l}\frac{\text{λ}}{\text{Δλ}}\text{=nN}\\ \text{Δλ=512}\text{.26}\text{nm-512}\text{.23=0}\text{.03}\text{nm}\end{array}$

The resolution is = $\frac{\text{λ}}{\text{Δλ}}$

    = $\frac{\text{512}}{0.03}$

     = $\text{17066}\text{.666}$

(b)

Interpretation Introduction

Interpretation:

The given resolution is close in nm is the closest line has to be resolved.

Concept introduction:

Resolution:

It is useful for separate two closely merged peaks

$\begin{array}{l}\frac{\text{λ}}{\text{Δλ}}\text{=nN}\\ \text{where,}\\ \text{λ}\text{is wavelength}\\ \text{n }\text{is the diffraction}\text{order}\\ \text{N}\text{is the number of grooves of the grating}\end{array}$

Answer to Problem 20.11P

The resolution is $\text{0}\text{.05}\text{nm}$

Explanation of Solution

Given,

Given,

Wavelengths is $\text{512}\text{.23}$

According to resolution of grating:

Resolution ${\text{10}}^{\text{-4}}$

$\begin{array}{l}\frac{\text{λ}}{\text{Δλ}}\text{=nN}\\ \text{Δλ=512}\text{.26}\text{nm-512}\text{.23=0}\text{.03}\text{nm}\end{array}$

The resolution is = $\frac{\text{λ}}{\text{Δλ}}=10000$

    = $\frac{\text{512}}{10000}$

     = $\text{0}\text{.05}\text{nm}$

(c)

Interpretation Introduction

Interpretation:

The fourth-order resolution of a grating has to be calculated.

Concept introduction:

Resolution of grating:

It is useful for separate two closely merged peaks

$\begin{array}{l}\frac{\text{λ}}{\text{Δλ}}\text{=nN}\\ \text{where,}\\ \text{λ}\text{is wavelength}\\ \text{n }\text{is the diffraction}\text{order}\\ \text{N}\text{is the number of grooves of the grating}\end{array}$

Answer to Problem 20.11P

The fourth-order resolution of a grating is $\text{ 5}{\text{.9×10}}^{\text{4}}$

Explanation of Solution

Given,

The fourth-order resolution of a grating is $\text{8}\text{.00}\text{cm}$ long is ruled at $\text{185lines/mm}$

According to resolution,

$\begin{array}{l}\text{=}\text{nN}\\ \text{=}\text{4 (80mm) (185}\text{lines/mm) = 5}{\text{.9×10}}^{\text{4}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The angle dispersion $\text{(}\Delta \varphi \text{)}$ between light rays with wavelengths has to be calculated.

Concept introduction:

Dispersion of grating:

It is useful to separate wavelengths differing with $\text{Δλ}$ through the difference in angle, $\text{(Δ}\varphi \text{)}$

$\frac{\text{Δφ}}{\text{Δλ}}\text{ = }\frac{\text{n}}{\text{dcosφ}}$

$\begin{array}{l}\text{where,}\\ \text{λ}\text{is wavelength}\\ \text{n }\text{is the diffraction}\text{order}\end{array}$

Answer to Problem 20.11P

The angle dispersion $\text{(}\Delta \varphi \text{)}$ between light rays with wavelengths is $\text{0}\text{.013°}$

Explanation of Solution

Given,

Wavelengths is $\text{512}\text{.23}\text{and 512}\text{.26}\text{nm}$

According to resolution of grating

The angular dispersion $\left(\text{Δφ}\right)$ between light rays with wavelengths of $\text{512}\text{.23}\text{and 512}\text{.26}\text{nm}$ nm for first-order diffraction (n = 1) and thirtieth-order diffraction from a grating with $\text{250 lines/mm and φ = 3}\text{.0°}$

$\begin{array}{l}\begin{array}{l}\text{Δλ = 0}\text{.03 nm, }\\ \text{d = 1/250 = 0}\text{.004 mm}\\ \end{array}\hfill \\ \frac{\text{Δφ}}{\text{Δλ}}\text{ = }\frac{\text{n}}{\text{dcosφ}}\hfill \end{array}$

$\begin{array}{l}\text{n = 1: Δφ = }\frac{\text{nΔλ}}{\text{dcosφ}}\hfill \\ \text{ =}\frac{\left(\text{0}\text{.03 nm}\right)}{\left[\text{0}\text{.004 mm·cos}\left(\text{3}\right)\right]}\hfill \\ \text{ = 7}{\text{.5 x 10}}^{\text{-6}}\text{ radians =0}\text{.0004°}\hfill \end{array}$

$\begin{array}{l}\text{n = 30: Δφ = }\frac{\text{nΔλ}}{\text{dcosφ}}\hfill \\ \text{ = 30}\left(\text{7}{\text{.5 x 10}}^{\text{-6}}\text{ radians}\right)\hfill \\ \text{ = 2}{\text{.25 x 10}}^{\text{-4}}\text{ radians =0}\text{.013°}\hfill \\ \hfill \end{array}$